vcaa 2006 mm cas 1
DESCRIPTION
vcaa examsTRANSCRIPT
SUPERVISOR TO ATTACH PROCESSING LABEL HERE
Figures
Words
STUDENT NUMBER Letter
Victorian Certifi cate of Education2006
MATHEMATICAL METHODS (CAS)Written examination 1
Friday 3 November 2006 Reading time: 9.00 am to 9.15 am (15 minutes) Writing time: 9.15 am to 10.15 am (1 hour)
QUESTION AND ANSWER BOOK
Structure of bookNumber ofquestions
Number of questionsto be answered
Number ofmarks
11 11 40
• Students are permitted to bring into the examination room: pens, pencils, highlighters, erasers, sharpeners, rulers.
• Students are NOT permitted to bring into the examination room: notes of any kind, blank sheets of paper, white out liquid/tape or a calculator of any type.
Materials supplied• Question and answer book of 10 pages, with a detachable sheet of miscellaneous formulas in the
centrefold.• Working space is provided throughout the book.
Instructions• Detach the formula sheet from the centre of this book during reading time.• Write your student number in the space provided above on this page.
• All written responses must be in English.
Students are NOT permitted to bring mobile phones and/or any other unauthorised electronic devices into the examination room.
© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006
2006 MATHMETH & MATHMETH(CAS) EXAM 1 2
This page is blank
3 2006 MATHMETH & MATHMETH(CAS) EXAM 1
TURN OVER
Question 1Let f (x) = x2 + 1 and g(x) = 2x + 1. Write down the rule of f (g(x)).
1 mark
Question 2For the function f : R → R, f (x) = 3e2x � 1,a. Þ nd the rule for the inverse function f −1
2 marks
b. Þ nd the domain of the inverse function f −1.
1 mark
InstructionsAnswer all questions in the spaces provided.A decimal approximation will not be accepted if an exact answer is required to a question.In questions where more than one mark is available, appropriate working must be shown.Unless otherwise indicated, the diagrams in this book are not drawn to scale.
2006 MATHMETH & MATHMETH(CAS) EXAM 1 4
Question 3a. Let f (x) = ecos (x). Find f ′(x)
1 mark
b. Let y = x tan (x). Evaluate dydx
when x = π6
.
3 marks
Question 4
For the function f :[ −π , π ] → R, f (x) = 5 23
cos x +
π
a. write down the amplitude and period of the function
2 marks
b. sketch the graph of the function f on the set of axes below. Label axes intercepts with their coordinates. Label endpoints of the graph with their coordinates.
3 marks
y
xO π– π
6
5
4
3
2
1
–1
–2
–3
–4
–5
–6
5 2006 MATHMETH & MATHMETH(CAS) EXAM 1
TURN OVER
Question 5Let X be a normally distributed random variable with a mean of 72 and a standard deviation of 8. Let Z be the standard normal random variable. Use the result that Pr(Z < 1) = 0.84, correct to two decimal places, to Þ nda. the probability that X is greater than 80
1 mark
b. the probability that 64 < X < 72
1 mark
c. the probability that X < 64 given that X < 72.
2 marks
2006 MATHMETH & MATHMETH(CAS) EXAM 1 6
Question 6The probability density function of a continuous random variable X is given by
f x
x x( ) =
≤ ≤
12
1 5
0 otherwise
a. Find Pr (X < 3).
2 marks
b. If Pr (X ≥ a) = 58 , Þ nd the value of a.
2 marks
7 2006 MATHMETH & MATHMETH(CAS) EXAM 1
TURN OVER
Question 7The graph of f : [�5, 1] → R where f (x) = x3 + 6x2 + 9x is as shown.
a. On the same set of axes sketch the graph of y = f x( ) .
2 marks
b. State the range of the function with rule y = f x( ) and domain [�5, 1].
1 mark
Question 8A normal to the graph of y x= has equation y = � 4x + a, where a is a real constant. Find the value of a.
4 marks
–3 Ox
y
2006 MATHMETH & MATHMETH(CAS) EXAM 1 8
Question 9A rectangle XYZW has two vertices, X and W, on the x-axis and the other two vertices, Y and Z, on the graph of y = 9 � 3x2, as shown in the diagram below. The coordinates of Z are (a, b) where a and b are positive real numbers.
a. Find the area, A, of rectangle XYZW in terms of a.
1 mark
b. Find the maximum value of A and the value of a for which this occurs.
3 marks
y
x
Y(–a, b) Z(a, b)
X WO
9 2006 MATHMETH & MATHMETH(CAS) EXAM 1
TURN OVER
Question 10Jo has either tea or coffee at morning break. If she has tea one morning, the probability she has tea the next morning is 0.4. If she has coffee one morning, the probability she has coffee the next morning is 0.3. Suppose she has coffee on a Monday morning. What is the probability that she has tea on the following Wednesday morning?
3 marks
CONTINUED OVER PAGE
2006 MATHMETH & MATHMETH(CAS) EXAM 1 10
Question 11Part of the graph of the function f : R → R, f (x) = �x2 + ax + 12 is shown below. If the shaded area is 45 square units, Þ nd the values of a, m and n where m and n are the x-axis intercepts of the graph of y = f (x).
5 marks
END OF QUESTION AND ANSWER BOOK
y
xn mO 3
12
MATHEMATICAL METHODS AND MATHEMATICAL METHODS (CAS)
Written examinations 1 and 2
FORMULA SHEET
Directions to students
Detach this formula sheet during reading time.
This formula sheet is provided for your reference.
© VICTORIAN CURRICULUM AND ASSESSMENT AUTHORITY 2006
MATH METH & MATH METH (CAS) 2
This page is blank
3 MATH METH & MATH METH (CAS)
END OF FORMULA SHEET
Mathematical Methods and Mathematical Methods CAS Formulas
Mensuration
area of a trapezium: 12
a b h+( ) volume of a pyramid: 13
Ah
curved surface area of a cylinder: 2π rh volume of a sphere: 43
3π r
volume of a cylinder: π r2h area of a triangle: 12
bc Asin
volume of a cone: 13
2π r h
Calculusddx
x nxn n( ) = −1
x dx
nx c nn n=
++ ≠ −+∫
11
11 ,
ddx
e aeax ax( ) = e dx a e cax ax= +∫1
ddx
xxelog ( )( ) = 1
1x
dx x ce= +∫ log
ddx
ax a axsin( ) cos( )( ) = sin( ) cos( )ax dx a ax c= − +∫1
ddx
ax a axcos( )( ) −= sin( ) cos( ) sin( )ax dx a ax c= +∫
1
ddx
ax aax
a axtan( )( )
( ) ==cos
sec ( )22
product rule: ddx
uv u dvdx
v dudx
( ) = + quotient rule: ddx
uv
v dudx
u dvdx
v
=−2
chain rule: dydx
dydu
dudx
= approximation: f x h f x h f x+( ) ≈ ( ) + ′ ( )
ProbabilityPr(A) = 1 – Pr(A′) Pr(A ∪ B) = Pr(A) + Pr(B) – Pr(A ∩ B)
Pr(A|B) =Pr
PrA B
B∩( )
( )mean: µ = E(X) variance: var(X) = σ 2 = E((X – µ)2) = E(X2) – µ2
probability distribution mean variance
discrete Pr(X = x) = p(x) µ = ∑ x p(x) σ 2 = ∑ (x – µ)2 p(x)
continuous Pr(a < X < b) = f x dxa
b( )∫ µ =
−∞
∞∫ x f x dx( ) σ µ2 2= −
−∞
∞∫ ( ) ( )x f x dx